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    <meta name="description" content="『算法-ACM 竞赛-疯子的算法总结』1 位运算（快速幂、快速乘）一、预备知识（补码，反码）计算机通过二进制表示整形数，比如 int 型 32 位有符号整形数：1 表示为：0000…..00001(共 32 位）-1 表示为：1111…..1111(共 32 位）补码计算法定义：非负数的补码是其原码本身；负数的补码是其绝对值的原码最高位符号位不变，其它位取反，再加 1。表示原因：计算机逻辑运算没有">
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            <h1 id="seo-header">『算法-ACM竞赛-疯子的算法总结』1  位运算（快速幂、快速乘）</h1>
            
            
              <div class="markdown-body">
                
                <h1 id="『算法-ACM-竞赛-疯子的算法总结』1-位运算（快速幂、快速乘）"><a href="#『算法-ACM-竞赛-疯子的算法总结』1-位运算（快速幂、快速乘）" class="headerlink" title="『算法-ACM 竞赛-疯子的算法总结』1 位运算（快速幂、快速乘）"></a>『算法-ACM 竞赛-疯子的算法总结』1 位运算（快速幂、快速乘）</h1><h2 id="一、预备知识（补码，反码）"><a href="#一、预备知识（补码，反码）" class="headerlink" title="一、预备知识（补码，反码）"></a>一、预备知识（补码，反码）</h2><p>计算机通过二进制表示整形数，比如 int 型 32 位有符号整形数：<br>1 表示为：0000…..00001(共 32 位）<br>-1 表示为：1111…..1111(共 32 位）<br>补码计算法定义：非负数的补码是其原码本身；<br>负数的补码是其绝对值的原码最高位符号位不变，其它位取反，再加 1。<br><strong>表示原因</strong>：计算机逻辑运算没有减法，-1+1 最高为溢出，剩余 0000000000（32 位）即为 0；<br>则有 a-b&#x3D;a+b 的（补码）；<br><strong>计算方式</strong>：<br>-1 表示原码为 100…….0001(32 位），最高位位符号位。<br>-1 的反码表示为：1111………110(32 位），除符号位按位取反。<br>-1 的补码表示为：1111………1111(32 位），反码+1。<br>正数的补码为自己本身。<br>例子：<br>100 的补码 ‭00000000000000000001100100‬<br>-30 的补码 11111111111111111111111100010‬<br>100+（-30）&#x3D;000000000000000000‭01000110‬<br>转换成 10 进制为 70；</p>
<h2 id="二、基本操作"><a href="#二、基本操作" class="headerlink" title="二、基本操作"></a>二、基本操作</h2><p><strong>1、按位与(&amp;)</strong></p>
<p>参加运算的两个数，换算为二进制(0、1)后，进行与运算。只有当相应位上的数都是 1 时，该位才取 1，否则该为为 0。</p>
<p>将 10 与-10 进行按位与(&amp;)运算：<br>|0000 0000 0000 1010<br>|——-<br>|<strong>1111 1111 1111 0110</strong><br>|<strong>0000 0000 0000 0010</strong><br>所以：10 &amp; -10 &#x3D; 0000 0000 0000 0010<br><strong>2、按位或(|)</strong></p>
<p>参加运算的两个数，换算为二进制(0、1)后，进行或运算。只要相应位上存在 1，那么该位就取 1，均不为 1，即为 0。</p>
<p>将 10 与-10 进行按位或(|)运算：</p>
<p>| <strong>0000 0000 0000 1010</strong><br>|——————<br>|<strong>1111 1111 1111 0110</strong><br>|<strong>1111 1111 1111 1110</strong></p>
<p>所以：10 | -10 &#x3D; 1111 1111 1111 1110<br><strong>3、按位异或(^)</strong></p>
<p>参加运算的两个数，换算为二进制(0、1)后，进行异或运算。只有当相应位上的数字不相同时，该为才取 1，若相同，即为 0。</p>
<p>将 10 与-10 进行按位异或(^)运算：</p>
<h2 id="0000-0000-0000-1010"><a href="#0000-0000-0000-1010" class="headerlink" title="| 0000 0000 0000 1010 |"></a>| 0000 0000 0000 1010 |</h2><p>|<strong>1111 1111 1111 0110</strong>|<br><strong>1111 1111 1111 1100</strong>|<br>所以：10 ^ -10 &#x3D; 1111 1111 1111 1100<br>可以看出，任何数与 0 异或，结果都是其本身。利用异或还可以实现一个很好的交换算法，用于交换两个数，算法如下：</p>
<p>a &#x3D; a ^ b;<br>b &#x3D; b ^ a;<br>a &#x3D; a ^ b;</p>
<p><strong>4、取反(~)</strong></p>
<p>参加运算的两个数，换算为二进制(0、1)后，进行取反运算。每个位上都取相反值，1 变成 0，0 变成 1。<br>对 10 进行取反(<del>)运算：<br>|0000 0000 0000 1010<br>|—————-|<br>|<strong>1111 1111 1111 0101</strong><br>所以：</del>10 &#x3D; 1111 1111 1111 0101<br><strong>5、左移(&lt;&lt;)</strong></p>
<p>参加运算的两个数，换算为二进制(0、1)后，进行左移运算，用来将一个数各二进制位全部向左移动若干位。</p>
<p>对 10 左移 2 位(就相当于在右边加 2 个 0)：</p>
<table>
<thead>
<tr>
<th>0000 0000 0000 1010</th>
</tr>
</thead>
<tbody><tr>
<td><strong>0000 0000 0010 1000</strong></td>
</tr>
<tr>
<td>所以：10 &lt;&lt; 2 &#x3D; 0000 0000 0010 1000 &#x3D; 40</td>
</tr>
<tr>
<td>注意，观察可以发现，左移一位的结果就是原值乘 2，左移两位的结果就是原值乘 4。</td>
</tr>
</tbody></table>
<p><strong>6、右移(&gt;&gt;)</strong></p>
<p>参加运算的两个数，换算为二进制(0、1)后，进行右移运算，用来将一个数各二进制位全部向右移动若干位。</p>
<p>对 10 右移 2 位(就相当于在左边加 2 个 0)：</p>
<table>
<thead>
<tr>
<th>0000 0000 0000 1010</th>
</tr>
</thead>
<tbody><tr>
<td><strong>0000 0000 0000 0010</strong></td>
</tr>
</tbody></table>
<p>所以：10 &gt;&gt; 2 &#x3D; 0000 0000 0000 0010 &#x3D; 2<br>注意，观察可以发现，右移一位的结果就是原值除 2，左移两位的结果就是原值除 4，注意哦，除了以后没有小数位的，都是取整。</p>
<h2 id="三、延伸操作"><a href="#三、延伸操作" class="headerlink" title="三、延伸操作"></a>三、延伸操作</h2><p><strong>1.快速幂（快速模幂）</strong><br>① 求 a^b:</p>
<pre><code class="hljs">int pow(int a, int k)  &#123;
    int ans = 1;
    while(k)  &#123;
        if(k &amp;1)  ans *= a;   //判断奇偶只用判断最后一位比取模快
        a *= a;
        k &gt;&gt;=1;		//比除法快多了
    &#125;
    return ans;
&#125;
</code></pre>
<p>② 求 a^b%p</p>
<figure class="highlight axapta"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br></pre></td><td class="code"><pre><code class="hljs axapta"><span class="hljs-built_in">int</span> pow_mod(<span class="hljs-built_in">int</span> a, <span class="hljs-built_in">int</span> k,<span class="hljs-built_in">int</span> <span class="hljs-keyword">mod</span>)  &#123;<br>     <span class="hljs-built_in">int</span> ans = <span class="hljs-number">1</span>%<span class="hljs-keyword">mod</span>;<br>     <span class="hljs-keyword">while</span>(k)  &#123;<br>         <span class="hljs-keyword">if</span>(k &amp;<span class="hljs-number">1</span>)  ans =(<span class="hljs-built_in">long</span> <span class="hljs-built_in">long</span>) ans*a%<span class="hljs-keyword">mod</span>;  <span class="hljs-comment">//防止在对P取模前溢出</span><br>         a = (<span class="hljs-built_in">long</span> <span class="hljs-built_in">long</span>)a*a%<span class="hljs-keyword">mod</span>;<br>         k &gt;&gt;=<span class="hljs-number">1</span>;  <span class="hljs-comment">//比除法快多了</span><br>     &#125;<br>     <span class="hljs-keyword">return</span> ans;<br> &#125;<br></code></pre></td></tr></table></figure>

<p>例题：BZOJ1008<br><strong>2.快速乘法</strong><br>方法 ①</p>
<figure class="highlight axapta"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br></pre></td><td class="code"><pre><code class="hljs axapta"><span class="hljs-built_in">long</span> <span class="hljs-built_in">long</span> quickMul(<span class="hljs-built_in">long</span> <span class="hljs-built_in">long</span> a,<span class="hljs-built_in">long</span> <span class="hljs-built_in">long</span> b,<span class="hljs-built_in">long</span> <span class="hljs-built_in">long</span> <span class="hljs-keyword">mod</span>)<br>&#123;<br>    <span class="hljs-built_in">long</span> <span class="hljs-built_in">long</span> ans=<span class="hljs-number">0</span>;<br>    <span class="hljs-keyword">while</span>(b)&#123;<br>        <span class="hljs-keyword">if</span>(b&amp;<span class="hljs-number">1</span>) ans=(ans+a)%<span class="hljs-keyword">mod</span>;<br>        a=(a+a)%<span class="hljs-keyword">mod</span>;  <span class="hljs-comment">//（计算机加法比乘法快，a+a比a*2快）</span><br>        b&gt;&gt;=<span class="hljs-number">1</span>;<br>    &#125;<br>    <span class="hljs-keyword">return</span> ans;<br>&#125;<br></code></pre></td></tr></table></figure>

<p>方法 ②：高效算法</p>
<figure class="highlight axapta"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br></pre></td><td class="code"><pre><code class="hljs axapta"><span class="hljs-built_in">long</span> <span class="hljs-built_in">long</span> quickMul(<span class="hljs-built_in">long</span> <span class="hljs-built_in">long</span> a,<span class="hljs-built_in">long</span> <span class="hljs-built_in">long</span> b,<span class="hljs-built_in">long</span> <span class="hljs-built_in">long</span> <span class="hljs-keyword">mod</span>)<br>&#123;<br>    a%=<span class="hljs-keyword">mod</span>;<br>    b%=<span class="hljs-keyword">mod</span>;<br>    <span class="hljs-built_in">long</span> <span class="hljs-built_in">long</span>  ans=<span class="hljs-number">0</span>;<br>    <span class="hljs-keyword">while</span>(b)&#123;<br>        <span class="hljs-keyword">if</span>(b&amp;<span class="hljs-number">1</span>)&#123;<br>            ans+=a;<br>            <span class="hljs-keyword">if</span>(ans&gt;=<span class="hljs-keyword">mod</span>)<br>               ans-=<span class="hljs-keyword">mod</span>;<br>        &#125;<br>        b&gt;&gt;=<span class="hljs-number">1</span>;<br>        a&lt;&lt;=<span class="hljs-number">1</span>;<br>        <span class="hljs-keyword">if</span>(a&gt;=<span class="hljs-keyword">mod</span>)  a-=<span class="hljs-keyword">mod</span>;<br>    &#125;<br>   <span class="hljs-keyword">return</span> ans;<br>&#125;<br></code></pre></td></tr></table></figure>

<p>方法 ③：使用 long double 优化版</p>
<figure class="highlight axapta"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br></pre></td><td class="code"><pre><code class="hljs axapta"><span class="hljs-built_in">long</span> <span class="hljs-built_in">long</span> quickMul(<span class="hljs-built_in">long</span> <span class="hljs-built_in">long</span> a,<span class="hljs-built_in">long</span> <span class="hljs-built_in">long</span> b,<span class="hljs-built_in">long</span> <span class="hljs-built_in">long</span> <span class="hljs-keyword">mod</span>)<br>&#123;<br>    a%=<span class="hljs-keyword">mod</span>;<br>    b%=<span class="hljs-keyword">mod</span>;<br>    <span class="hljs-built_in">long</span> <span class="hljs-built_in">long</span> c=(<span class="hljs-built_in">long</span> <span class="hljs-built_in">double</span>) a*b/<span class="hljs-keyword">mod</span>;<br>    <span class="hljs-built_in">long</span> <span class="hljs-built_in">long</span> ans=a*b-c*<span class="hljs-keyword">mod</span>;<br>    <span class="hljs-keyword">if</span>(ans&lt;<span class="hljs-number">0</span>) ans+=<span class="hljs-keyword">mod</span>;<br>    <span class="hljs-keyword">else</span> <span class="hljs-keyword">if</span>(ans&gt;=<span class="hljs-keyword">mod</span>) ans-=<span class="hljs-keyword">mod</span>;<br>    <span class="hljs-keyword">return</span> ans<br>  &#125;<br></code></pre></td></tr></table></figure>

<p><strong>在这里仅提到部分操作，在 ACM 学习中，还有更多的操作可以用位运算。</strong></p>

                
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